$f: [0,1] \to [0,1]$ is injective & $ f(2x-f(x))=x$ for $x \in [0,1]$ then $f(x)=x $

Question

Let $f: [0,1] \to [0,1]$ be an injective function such that $ f(2x-f(x))=x$ for $x \in [0,1]$. I have to prove that $f(x)=x, x\in [0,1]$.
My attempt:
let $$g(x)=2x-f(x)$$ then $$f(g(x))=x \implies f$$ has an inverse, $$f^{-1}=2x-f(x) $$,
and I have no clue on how to proceed from here, the solution at the back of the back suggests defining a sequence $(x_{n})$ recursively as, $(x_{n})=f(x_{n-1})$ but I couldn't understand it further.
Help of any sort (hints or full solution, whatever seems possible) would be greatly appreciated

Answer 1

$f(2x -f(x)) = x, y=f(x), f(2y-f(y))=y$ implies that $f(2f(x)-f^2(x))=f(x)$

f injective $2f(x)-f^2(x)= x, 2x=4f(x)- 2f^2(x), 2x-f(x) = 3f(x)-2f^2(x)$ this implies $f(2x-f(x))= f(3f(x)-2f^2(x))=x$ f injective

$3f(x)-2f^2(x)=x$, since $2f(x)-f^2(x)= x, f(x)=x$ $3f(x)-2f^2(x)-2(2f(x)-f^2(x))=-f(x)=x-2x$, thus f(x)=x

Answer 2

I use what you have done. You have $f^{-1}(x)=2x-f(x)$. Now replace $x$ by $f(x)$, you get $f(f(x))-2f(x)+x=0$. Now fix $x$, and define $x_n$ by $x_1=x$ and $x_{n+1}=f(x_n)$. Replacing $x$ by $x_n$, you get $f(f(x_n))-2f(x_n)+x_n=0$, hence $x_{n+2}-2x_{n+1}+x_n=0$, or $x_{n+2 }-x_{n+1}=x_{n+1}-x_n$. This show that $x_{n+1}-x_n$ is a constant $c$, and $x_n=(n-1)c+x_1$. Now $c$ is zero because $x_n\in [0,1]$ for all $n$, and we get $x_2=f(x)=x_1=x$.